Answer: C) 35 and 11 years Explanation: Let the age of sushma be x and the age of her son is yThen five years before x-5=5(y-5) ...(1)Five years hence x+5 = 3(y+5)-8 .....(2) By soving (1) & (2), we get5y - 15 = 3y + 7y = 11 => x = 35 Therefore, the age of Sushma = 35 and her son = 11.

Answer: C) Both A & B gives result Explanation: a and b give, profit after 3 years = Rs.(3/8 x 22000) = Rs.8250. From c also, profit after 3 years = Rs. (2750 x 3) = Rs. 8250. ∴ P's share = Rs.(8250 x 5/11) = Rs. 3750. Thus, (either C is redundant) or (a and b are redundant).

Answer: D) Data is not sufficient Explanation: Let the sum be Rs. p, rate be R% p.a. and time be T years. Then,  P×T×(R+2)100-P×T×R100=108  ⇒2PT= 10800  ...... (1)  And,   P×R×(T+2)100-P×R×T100=180   ⇒2PR = 18000 ......(2) Clearly, from (1) and (2), we cannot find the value of pSo, the data is not sufficient.

Answer: A) Rs. 404.80 Explanation: Let C1 be the cost price of the first article and C2 be the cost price of the second article.Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%. We know, C1 + C2 = 600.The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2. The total selling price of the first and second article = 1.22C1 + 0.92C2. As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2. Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get1.22C1 + 0.92(600 - C1) = 600or 1.22C1 - 0.92C1 = 600 - 0.92*600or 0.3C1 = 0.08*600 = 48or C1 = 48/(0.3) = 160.If C1 = 160, then C2 = 600 - 160 = 440.The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.

Answer: B) Rs.12,000 Explanation: We have P = 8748, R = -10 and n = 3Therefore, the purchase price of the machine= P1+R100n= 87481-101003= 8748 x (100/90) x (100/90) x (100/90)= 12,000.   Therefore, the purchase price of the machine was Rs. 12,000.

Answer: C) 175 Explanation: Given that,  working days = 5  working hours = 8  A man get rupess per hour is Rs.2.40  So in one day the man get total rupees is 2.40 x 8 = 19.2  So in 5 days week the man get total rupees is 19.2 x 5 = 96  So in 4 week the man get total rupees is 96 x 4 = 384  So the man worked for = 160hours in 4 weeks    But given that the man earned Rs.432    Hence remaning money is (432-384 = 48)which is earn by doing overtime work   Overtime hours  = 48/3.20 = 15 So total worked hours is = 15 + 160 = 175.

Answer: A) 45 metric tonnes Explanation: 2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton  i.e. 3 engines of latter type consumes 1 ton for one hour Hence 9 engines consumes 3 tons for one hour For 15 hours it is 15 x 3 = 45 metric tonnes.

Answer: B) 13 + 1/3 days Explanation: C alone can finish the work in 40 days.As given C does half as much work as A and B together => (A + B) can do it in 20 days(A + B)s 1 days wok = 1/20.A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]B's 1 days work = (1/20) x (2/3) = 1/30(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40 All the three together will finish it in 40/3 = 13 and 1/3 days.